Answer:
The frequency of the phonograph record is 0.2 Hz
Explanation:
The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period
The given parameters of the phonograph record are;
The radius of the record = 0.15 m
The number of times the phonograph record rotates, n = 18 times
The time it takes the phonograph record to rotate the 18 times, t = 90 seconds
The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)
∴ The frequency of the phonograph record, f = n/t = 18/(90 s) = 0.2 Hz
The frequency of the phonograph record = 0.2 Hz.
Use the formula F = (9x10^9 Q * q) / r^2
Message me if you need help.
Answer:
A hypothesis is what you think will happen.
A conclusion is the results of an experiment summarized.
Hope this helps.
Answer:
Energy dissipated = 13.453 Joules
Explanation:
In order to solve this problem, we first compute the gravitational potential energy the child has, and then find the kinetic energy at the lowest position.
The gravitational potential energy (relative to lowest position) is found as follows:
G.P.E = mass * gravity * height
Where, Height = 2 - 2 * Cos(34°)
Height = 0.3193 meters
G.P.E = 30 * 9.8 * 0.3193
G.P.E = 93.874 J
Kinetic energy:
K.E = 0.5 * mass * velocity^2
K.E = 0.5 * 30 * 2.31547^2
K.E = 80.421 J
Energy dissipated = G.P.E - K.E
Energy dissipated = 93.874 - 80.421
<u>Energy dissipated = 13.453 J</u>
Answer:
1. spring potential energy and gravitational potential energy
2. 490 joules
3. 75 joules
4. 650 joules
5. thermal energy